Question

Find the absolute maximum and the absolute minimum values of `f(x) = (x + 1)/(x^2 + x + 9)` on the interval `[ 0,∞)`?

Answer 1

On the interval `[0,oo)`, we have absolute maxima at `x=2`, where `f(x)=1/5`

Explanation:

At absolute maximum or minimum derivative of `f(x)=(x+1)/(x^2+x+9)` will be zero. We have a maximum when second derivative is negative.

`(df)/(dx)=((x^2+x+9)xx1-(x+1)(2x+1))/(x^2+x+9)^2`

= `(x^2+x+9-2x^2-3x-1)/(x^2+x+9)^2`

= `-(x^2+2x-8)/(x^2+x+9)^2`

= `-((x+4)(x-2))/(x^2+x+9)^2`

as such we have extrema at `x=-4` and `x=2` and in the interval we have just one extrema at `x=2` and

`(d^2f)/(dx^2)=-((x^2+x+9)^2(2x+2)-2(x^2+x+9)(2x+1)(x^2+2x-8))/(x^2+x+9)^4`

= `-((x^2+x+9)((x^2+x+9)(2x+2)-(4x+2)(x^2+2x-8)))/(x^2+x+9)^4`

= `-(2x^3+4x^2+20x+18-(4x^3+10x^2-28x-16))/(x^2+x+9)^3`

= `(2x^3+6x^2-48x-34)/(x^2+x+9)^3`

at `x=2`, we have `(d^2f)/(dx^2)=-2/75`

and at `x=-4`, `(d^2f)/(dx^2)=2/147`

Hence we have an absolute maxima at `x=2` and `f(2)=3/15=1/5`

And as degree of denominator is higher, while at `x=0`, `f(0)=1/9`, as `x->oo`, `f(x)->0`

graph{y-(x+1)/(x^2+x+9)=0 [-10, 10, -0.4, 0.4]}

graph{y-(x+1)/(x^2+x+9)=0 [-200, 200, -0.4, 0.4]}